AES / CFB8 IV尺寸 [英] AES/CFB8 IV size

问题描述

AFAIK，CFB8模式的块大小为1byte。所以我可以诱导IV也是1个字节长度。
然而，当我做一个测试传递同样iv只有1个字节到通用crypto创建函数加密和解密函数，加密和解密的消息不匹配。

AFAIK, CFB8 mode has block size of 1byte. So I can induce that IV is also 1byte length. However, when I do a test passing same iv of just 1 byte into common crypto create function for encrypt and decrypt function, encrypted and decrypted message mismatch.

所以我认为API应该使用超过1个字节作为IV。我想知道为什么？

So I think that the API should have taken more than 1 byte to use as IV. I would like to know why? Any thing wrong with my understanding?

CCCryptorStatus result = CCCryptorCreateWithMode(operation,
                                                 kCCModeCFB8,
                                                 kCCAlgorithmAES128,
                                                 ccNoPadding,
                                                 iv.bytes,
                                                 key.bytes,
                                                 key.length,
                                                 NULL,
                                                 0,
                                                 0,
                                                 0,
                                                 &_cryptor);

if (result == kCCSuccess)
    result = CCCryptorUpdate(_cryptor,
                             data.bytes,
                             data.length,
                             cipherData.mutableBytes,
                             cipherData.length,
                             &outLength);

if (result == kCCSuccess)
    result = CCCryptorFinal(_cryptor,
                            cipherData.mutableBytes,
                            cipherData.length,
                            &outLength);

if (result == kCCSuccess)
    result = CCCryptorRelease(_cryptor);

推荐答案

它只是每1个字节重新同步。
IV实际上是16字节（对于AES）。

It doesn't have block size of 1 byte, it is just resynchronized each 1 byte. The IV is actually 16 bytes (for AES).

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